SOLUTION: find the canonical form and determine the nature of conics 8x^2-12xy+17y^2+16x-12y+3=0

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Question 1147397: find the canonical form and determine the nature of conics
8x^2-12xy+17y^2+16x-12y+3=0
Answer by KMST(5398) About Me  (Show Source):
You can put this solution on YOUR website!
A quadratic equation of the form Ax%5E2%2BBxy%2BCy%5E2%2BDx%2BEy%2BF=0 , like 8x%5E2-12xy%2B17y%5E2%2B16x-12y%2B3=0 , could represent a circle, ellipse, hyperbola, parabola.
In special cases it could represent a point, a line, a pair of lines, or no point that could exist, depending on the values of the coefficients.
The value B%5E2-4AC , called the discriminant, suggests parabola, ellipse, or hyperbola if it is zero, negative, or positive respectively.
In the case of 8x%5E2-12xy%2B17y%5E2%2B16x-12y%2B3=0 , with system%28A=8%2CB=-12%2CC=17%29 ,
B%5E2-4AC=%28-12%29%5E2-4%2A8%2A17=144-544=-400%3C0 suggest that the equation represents an ellipse.
If there was no term in xy , the axes of symmetry of the ellipse would be parallel to the x- an y-axes.
If there is term in xy , one of the axes of symmetry will be at a positive angle alpha to the positive x-axis such that 0%5Eo%3Calpha%3C90%5Eo .
Rotating the coordinate axes counterclockwise any angle alpha such that 0%5Eo%3Calpha%3C90%5Eo ,
a point P%28x%2Cy%29 would be called the point P%28u%2Cv%29 with coordinates referencing the new axes,
with u=x%2Acos%28alpha%29%2By%2Asin%28alpha%29 and v=-x%2Asin%28alpha%29%2By%2Acos%28alpha%29 .
For the reverse conversion, highlight%28x=u%2Acos%28alpha%29-v%2Asin%28alpha%29%29 and highlight%28y=u%2Asin%28alpha%29%2Bv%2Acos%28alpha%29%29.

The points represented by equation Ax%5E2%2BBxy%2BCy%5E2%2BDx%2BEy%2BF=0 would be represented in the new set of axes
by a new equation in u and v , with new coefficients. Ju%5E2%2BKuv%2BLv%5E2%2BMu%2BNv%2BF=0 .
Substituting highlight%28u%2Acos%28alpha%29-v%2Asin%28alpha%29%29 forx and highlight%28u%2Asin%28alpha%29%2Bv%2Acos%28alpha%29%29 for y into the original equation,
we would get the equation Ju%5E2%2BKuv%2BLv%5E2%2BMu%2BNv%2BF=0 .
Expressions to calculate the new coefficients can be found, including we must find a value for alpha that makes K=0
B%28cos%5E2%28alpha%29-sin%5E2%28alpha%29%29%2B2%28C-A%29sin%28alpha%29cos%28alpha%29=0 --> B%28cos%5E2%28alpha%29-sin%5E2%28alpha%29%29=2%28A-C%29sin%28alpha%29cos%28alpha%29 --> B%2F%28A-C%29%22=%222sin%28alpha%29cos%28alpha%29%2F%28cos%5E2%28alpha%29-sin%5E2%28alpha%29%29
That can be converted into highlight%28B%2F%28A-C%29=sin%282alpha%29%2Fcos%282alpha%29=+tan%282alpha%29%29 by using the trigonometric identities cos%282alpha%29=+cos%5E2%28alpha%29-sin%5E2%28alpha%29 and sin%282alpha%29=+sin%28alpha%29cos%28alpha%29
From there the values for 2alpha alpha and its trigonometric functions can be found.
With those values the other coefficients can be calculated as
J=A%2Acos%5E2%28alpha%29+%2BB%2Asin%28alpha%29cos%28alpha%29%2BC%2Asin%5E2%28alpha%29+ ,
L=+A%2Asin%5E2%28alpha%29+-B%2Asin%28alpha%29cos%28alpha%29%2BC%2Acos%5E2%28alpha%29 ,
M=+D%2Acos%28alpha%29%2BE%2Asin%28alpha%29 , and N=-D%2Asin%28alpha%29%2BE%2Acos%28alpha%29 .
After that, the canonical form for the equation in u and v can be found.

FINDING alpha AND ITS TRIGONOMETRIC FUNCTIONS:
As a quadratic equation of the form Ax%5E2%2BBxy%2BCy%5E2%2BDx%2BEy%2BF=0 , 8x%5E2-12xy%2B17y%5E2%2B16x-12y%2B3=0 has system%28A=8%2CB=-12%2CC=17%2CD=16%2CE=12%2CF=3%29
B%2F%28A-C%29%22=%22sin%282alpha%29%2Fcos%282alpha%29%22=%22tan%282alpha%29%22=%22%28-12%29%2F%288-17%29%22=%22%28-12%29%2F%28-9%29%22=%2212%2F9%22=%224%2F3%29
A calculator could provide a good approximation of the value of 2alpha=53.13%5Eo(rounded) and that value could be used to get approximated values for sin%282alpha%29 , cos%282alpha%29 , alpha , sin%28alpha%29 , and cos%28alpha%29 .
Alternatively, exact values can be found using trigonometric identities.
Because 0%5Eo%3Calpha%3C90%5Eo we know that 0%5Eo%3C2alpha%3C180%5Eo}, but because tan%282alpha%29=4%2F3%3E0 we conclude 0%5Eo%3C2alpha%3C90%5Eo as well, and al trigonometric functions will be positive.
From sin%282alpha%29%2Fcos%282alpha%29=4%2F3 <--> sin%282alpha%29=%284%2F3%29cos%282alpha%29 and trigonometric identity sin%5E2%282alpha%29%2Bcos%5E2%282alpha%29=1 we get
%28%284%2F3%29cos%282alpha%29%29%5E2%2Bcos%5E2%282alpha%29=1 --> %2816%2F9%29cos%5E2%282alpha%29%2Bcos%5E2%282alpha%29=1 --> %2816%2F9%2B1%29cos%5E2%282alpha%29=1 --> %2825%2F9%29cos%5E2%282alpha%29=1 --> cos%5E2%282alpha%29=9%2F25 --> cos%282alpha%29=sqrt%289%2F25%29=3%2F5=0.6 and sin%282alpha%29=%284%2F3%29cos%282alpha%29=%284%2F3%29%283%2F5%29=4%2F5=0.8 .
From cos%282alpha%29=3%2F5 and the trig identities for half angles we can find sin%5E2%28alpha%29 , sin%28alpha%29 , cos%5E2%28alpha%29 , cos%28alpha%29 , as
--> sin%28alpha%29=sqrt%281%2F5%29=sqrt%285%2F25%29=highlight%28sqrt%285%29%2F5%29
--> cos%28alpha%29=sqrt%284%2F5%29=highlight%282sqrt%285%29%2F5%29
We could also get the product highlight%28sin%28alpha%29cos%28alpha%29=2%2F5%29 from 4%2F5=sin%282alpha%29=2sin%28alpha%29cos%28alpha%29 --> %281%2F2%29%282sin%28alpha%29cos%28alpha%29%29=%281%2F2%29%284%2F5%29=2%2F5 .

CALCULATING J , L , M AND N :
Now we can substitute the highlighted values above into the equations to calculate J , L , M , and N :
J=A%2Acos%5E2%28alpha%29+%2BB%2Asin%28alpha%29cos%28alpha%29%2BC%2Asin%5E2%28alpha%29=8%2A%284%2F5%29%2B%28-12%29%282%2F5%29%2B17%2A%281%2F5%29=32%2F5%2B17%2F5-24%2F5=25%2F5=5 ,
%28rounded%29 , and
%28rounded%29 .

FINDING THE CANONICAL FORM AND THE NATURE OF THIS CONIC:
For now, substituting the values found for J and L , we can write the equation of the conic in terms of u and v as:
5u%5E2%2B20v%5E2%2BMu%2BNv%2B3=0 ,
and we can and need to transform it into something of the form %28u-h%29%5E2%2Fp%5E2%2B%28v-k%29%5E2%2Fq%5E2=1 which would be equivalent to
%28%28u-h%29%2Fp%29%5E2%2B%28%28v-k%29+%2Fq%29%5E2=1 .
That is the equation of a circle of radius 1 , stretched in the u and v directions by factors p and q , centered in point (h,k) .
In other words, that represents an ellipse centered in (h,k) .
To get that transformation we need to form the squares %28u-h%29%5E2=u-2uh%2Bh%5E2 and %2B%28v-k%29%5E2=v-2vk%2Bk%5E2 from 5u%5E2%2BMu and 20v%5E2+%2BNv respectively .
5u%5E2%2B20v%5E2%2BMu%2BNv%2B3=0 --> 5u%5E2%2BMu%2B20v%5E2%2BNv=-3 --> -->

M%2F10=4sqrt%285%29%2F10=highlight%280.4sqrt%285%29%29=0.89443%28rounded%29 ,
and N%2F40=-8sqrt%285%29%2F40=highlight%28-0.2sqrt%285%29%29=-0.447215%28rounded%29 .
%28M%2F10%29%5E2=%280.4sqrt%285%29%29%5E2=0.16%2A5=0.8
%28N%2F40%29%5E2=%28-0.2sqrt%285%29%29%5E2=0.04%2A5=0.2
Then, the right hand side term is -3%2B5%28M%2F10%29%5E2%2B20%28N%2F40%29%5E2=-3%2B5%2A%280.4sqrt%285%29%29%5E2%2B20%2A%280.5sqrt%285%29%29%5E2=-3%2B5%2A0.16%2A5%2B20%2A0.04%2A5=-3%2B4%2B4=5 , and the equation is
highlight%285%28u%2B0.4sqrt%285%29%29%5E2%2B20%28v%2B%28-0.2sqrt%282%29%29%29%5E2=5%29
From there, we continue:
5%28u%2B0.4sqrt%285%29%29%5E2%2B20%28v-0.2sqrt%285%29%29%5E2=5 --> %28u%2B0.4sqrt%285%29%29%5E2%2B4%28v-0.2sqrt%282%29%29%5E2=1 -->
%28u-%28-0.4sqrt%285%29%29%29%5E2%2F1%5E2%22%2B%22%28v-0.2sqrt%285%29%29%5E2%2F0.5%5E2%22=%221
The center of the ellipse is at %22%28%22-0.4sqrt%285%29%22%2C%220.2sqrt%285%29%22%29%22 .
The numbers squared in the denominators are the semi-major and semi-minor axes of the ellipse.
The greater one, 1 , is the semi-major axis, and the other I the semi-minor axis
The extreme values for u happen when v-0.2sqrt%285%29=0 , and then
%28u%2B0.4sqrt%285%29%29%5E2=1 --> u%2B0.4sqrt%285%29=%22+%22%2B-+1 --> u=-0.4sqrt%285%29+%2B-+1 .
The y-coordinate of the center and the extremes show the vertices are at a distance of 1 from the center .
Using the equations highlight%28x=u%2Acos%28alpha%29-v%2Asin%28alpha%29%29 and highlight%28y=u%2Asin%28alpha%29%2Bv%2Acos%28alpha%29%29 , we can find the coordinates of the center and vertices of the ellipse.
For the center, x=%28-0.4sqrt%285%29%29%2A%282sqrt%285%29%2F5%29-0.2sqrt%285%29%2A%28sqrt%285%29%2F5%29=
-0.8%2A5%2F5-0.04sqrt%2825%29=-0.8-0.04%2A5=-0.8-0.2=-1
y=-0.4sqrt%285%29%2A%28sqrt%285%29%2F5%29%2B0.2sqrt%285%29%2A%282sqrt%285%29%2F5%29=
-0.4%2A5%2F5%2B0.2%2A2%2A5%2F5=-0.4%2B0.4=0